Site Statistics

Some ambiguous points on Spontaneous Symmetry Breaking (SSB)?

+ 3 like- 0 dislike

56 views

Almost in every textbook of condensed matter physics, the standard description of SSB could be formulated as follows:

Consider the lattice Heisenberg model in an external magnetic field $H=\sum_{ij}J_{ij}\mathbf{S}_i\cdot\mathbf{S}_j+hS_z$, where $h$ is the magnitude of magnetic field and $S_z=\sum_iS_i^z$. Now the average magnetization per site is a function of both magnetic field $h$ and number of lattice sites $N$, say $m\equiv \sum_i\left \langle S_i^z \right \rangle/N=m(N,h)$, where $\left \langle S_i^z \right \rangle\equiv tr(\hat{\rho }S_i^z)$ with $\hat{\rho }=e^{-\beta H}/tr(e^{-\beta H})$ the density operator. Then if $$\lim_{h\rightarrow 0}\lim_{N\rightarrow \infty }m(N,h)\neq 0$$, we say the system has SSB at temperature $T$. Now I get some questions:

(1)We know at finite $N$ and zero $h$, $m(N,h=0)=0$ due to spin-rotation symmetry. But there is no reason for that $$\lim_{h\rightarrow 0}m(N,h)=m(N,h=0)—[1]$$, right? Since the function $m(N,h)$ may not be continuous at $h=0$, from the math viewpoint.

(3)If Eq.[1] is wrong, say $\lim_{h\rightarrow 0}m(N,h)\neq m(N,h=0)$ and hence $\lim_{h\rightarrow 0}m(N,h)\neq0$, then what about $$\lim_{N\rightarrow \infty }\lim_{h\rightarrow 0}m(N,h)?$$ And why don't we use this identity to define SSB?

What you write in point (2) is correct, since $m(N,h)$ is an analytic function of $h$ (and $\beta$) when $N$ is finite (this should be clear by looking at how $m(N,h)$ depends of $h$). This is actually the reason one has to take the thermodynamic limit first.

@K-boy: You can see that at finite $N$, the partition function (or the expectation value) is just the sum of a finite number of exponentials. This is therefore analytic, and the limit $h\to 0$ is continuous.

@ Adam, good point. But from the math viewpoint, even though the number of summation terms is finite, there is still possibility that some of the terms are singularity. For example, the partition function $$Z=\sum_{\alpha=1}^{d} e^{-\beta E_\alpha(h)}$$ with $d=2^N$ the dimension of the Hilbert space of $N$ spin-1/2 system and $E_\alpha(h)$ the eigenenergy of $H$, if there exist some $E_\alpha(h)$ which are not continuous functions of $h$, then the whole story may be not continuous on $h$. Is this possible?

@K-boy: Usually, one expects that the eigenvalues are analytical functions of $h$ if the Hilbert space is finite (though I don't have a proof of that). Of course, you can always invent a model where that's not the case, but that's not typical. Where you are right is that even though the system is finite, and that the eigenvalues are all analytical, the partition function can be non-analytical in $h$ in the limit $\beta\to \infty$ (you can show that with only one quantum spin). This corresponds to a kind of quantum phase transition associated with a level-crossing.

Your comment on this question:

To answer, leave an answer instead. Comments are usually for non-answers.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
To alert a user, please use the "@" command and remove spaces from the username, example, the user "John Doe" should be pinged as "@JohnDoe", while the user "Johndoe" should be pinged as "@Johndoe". The post author is always automatically pinged (unless you are the post author).
Please consult the FAQ for as to how to format your post.

Live preview (may slow down editor)Preview

Live Preview

Preview

Your name to display (optional):

Email me at this address if a comment is added after mine:Email me if a comment is added after mine

Privacy: Your email address will only be used for sending these notifications.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.

Live preview (may slow down editor)Preview

Live Preview

Preview

Your name to display (optional):

Email me at this address if my answer is selected or commented on:Email me if my answer is selected or commented on

Privacy: Your email address will only be used for sending these notifications.

Anti-spam verification:

If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ys$\varnothing$csOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).